Query Processing: Overview, Cost Measures, and Operations

1. Chapter 13: Query Processing

2. Chapter 13: Query Processing • Overview • Measures of Query Cost • Selection Operation • Sorting • Join Operation • Other Operations • Evaluation of Expressions

3. Basic Steps in Query Processing 1. Parsing and translation 2. Optimization 3. Evaluation

4. Basic Steps in Query Processing (Cont.) • Parsing and translation • translate the query into its internal form. This is then translated into relational algebra. • Parser checks syntax, verifies relations • Evaluation • The query-execution engine takes a query-evaluation plan, executes that plan, and returns the answers to the query.

5. Basic Steps in Query Processing : Optimization • A relational algebra expression may have many equivalent expressions • E.g., balance2500(balance(account)) is equivalent to balance(balance2500(account)) • Each relational algebra operation can be evaluated using one of several different algorithms • Correspondingly, a relational-algebra expression can be evaluated in many ways. • Annotated expression specifying detailed evaluation strategy is called an evaluation-plan. • E.g., can use an index on balance to find accounts with balance < 2500, • or can perform complete relation scan and discard accounts with balance  2500

6. Basic Steps: Optimization (Cont.) • Query Optimization: Amongst all equivalent evaluation plans choose the one with lowest cost. • Cost is estimated using statistical information from the database catalog • e.g. number of tuples in each relation, size of tuples, etc. • In this chapter we study • How to measure query costs • Algorithms for evaluating relational algebra operations • How to combine algorithms for individual operations in order to evaluate a complete expression • In Chapter 14 • We study how to optimize queries, that is, how to find an evaluation plan with lowest estimated cost

7. Measures of Query Cost • Cost is generally measured as total elapsed time for answering query • Many factors contribute to time cost • disk accesses, CPU, or even network communication • Typically disk access is the predominant cost, and is also relatively easy to estimate. Measured by taking into account • Number of seeks * average-seek-cost • Number of blocks read * average-block-read-cost • Number of blocks written * average-block-write-cost • Cost to write a block is greater than cost to read a block • data is read back after being written to ensure that the write was successful

8. Measures of Query Cost (Cont.) • For simplicity we just use the number of block transfers from disk and the number of seeks as the cost measures • tT – time to transfer one block • tS – time for one seek • Cost for b block transfers plus S seeksb * tT + S * tS • We ignore CPU costs for simplicity • Real systems do take CPU cost into account • We do not include cost to writing output to disk in our cost formulae • Several algorithms can reduce disk IO by using extra buffer space • Amount of real memory available to buffer depends on other concurrent queries and OS processes, known only during execution • We often use worst case estimates, assuming only the minimum amount of memory needed for the operation is available • Required data may be buffer resident already, avoiding disk I/O • But hard to take into account for cost estimation

9. Selection Operation • File scan – search algorithms that locate and retrieve records that fulfill a selection condition. • Algorithm A1 (linear search). Scan each file block and test all records to see whether they satisfy the selection condition. • Cost estimate = br block transfers + 1 seek • br denotes number of blocks containing records from relation r • If selection is on a key attribute, can stop on finding record • cost = (br /2) block transfers + 1 seek • Linear search can be applied regardless of • selection condition or • ordering of records in the file, or • availability of indices

10. Selection Operation (Cont.) • A2 (binary search). Applicable if selection is an equality comparison on the attribute on which file is ordered. • Assume that the blocks of a relation are stored contiguously • Cost estimate (number of disk blocks to be scanned): • cost of locating the first tuple by a binary search on the blocks • log2(br) * (tT + tS) • If there are multiple records satisfying selection • Add transfer cost of the number of blocks containing records that satisfy selection condition • Will see how to estimate this cost in Chapter 14

11. Selections Using Indices • Index scan – search algorithms that use an index • selection condition must be on search-key of index. • A3 (primary index on candidate key, equality). Retrieve a single record that satisfies the corresponding equality condition • Cost = (hi+ 1) * (tT + tS) height of the B+ tree: hi • A4 (primary index on nonkey, equality) Retrieve multiple records. • Records will be on consecutive blocks • Let b = number of blocks containing matching records • Cost = hi * (tT + tS)+ tS + tT * b • A5 (equality on search-key of secondary index). • Retrieve a single record if the search-key is a candidate key • Cost = (hi+ 1) * (tT + tS) • Retrieve multiple records if search-key is not a candidate key • each of n matching records may be on a different block • Cost = (hi+ n) * (tT + tS) • Can be very expensive!

12. Selections Involving Comparisons • Can implement selections of the form AV (r) or A  V(r) by using • a linear file scan or binary search, • or by using indices in the following ways: • A6 (primary index, comparison). (Relation is sorted on A) • For A  V(r) use index to find first tuple  v and scan relation sequentially from there • For AV (r) just scan relation sequentially till first tuple > v; do not use index • A7 (secondary index, comparison). • For A  V(r) use index to find first index entry  v and scan index sequentially from there, to find pointers to records. • For AV (r) just scan leaf pages of index finding pointers to records, till first entry > v • In either case, retrieve records that are pointed to • requires an I/O for each record • Linear file scan may be cheaper

13. Implementation of Complex Selections • Conjunction: 1 2. . .  n(r) • A8 (conjunctive selection using one index). • Select a combination of i and algorithms A1 through A7 that results in the least cost for i (r). • Test other conditions on tuple after fetching it into memory buffer. • A9 (conjunctive selection using multiple-key index). • Use appropriate composite (multiple-key) index if available. • A10 (conjunctive selection by intersection of identifiers). • Requires indices with record pointers. • Use corresponding index for each condition, and take intersection of all the obtained sets of record pointers. • Then fetch records from file • If some conditions do not have appropriate indices, apply test in memory.

14. Algorithms for Complex Selections • Disjunction:1 2 . . .  n (r). • A11 (disjunctive selection by union of identifiers). • Applicable if all conditions have available indices. • Otherwise use linear scan. • Use corresponding index for each condition, and take union of all the obtained sets of record pointers. • Then fetch records from file • Negation: (r) • Use linear scan on file • If very few records satisfy , and an index is applicable to  • Find satisfying records using index and fetch from file

15. Sorting • We may build an index on the relation, and then use the index to read the relation in sorted order. May lead to one disk block access for each tuple. • For relations that fit in memory, techniques like quicksort can be used. For relations that don’t fit in memory, external sort-merge is a good choice.

16. External Sort-Merge Let M denote memory size (in pages). • Create sortedruns. (run: sorted subfile – is stored at disk) Let i be 0 initially.Repeatedly do the following till the end of the relation: (a) Read M blocks of relation into memory (b) Sort the in-memory blocks (c) Write sorted data to run Ri; increment i.Let the final value of i be N Merge the runs (next slide)….. N – the number of runs

17. External Sort-Merge (Cont.) • Merge the runs (N-way merge). We assume (for now) that N < M. • Use N blocks of memory to buffer input runs, and 1 block to buffer output. Read the first block of each run into its buffer page • repeat • Select the first record (in sort order) among all buffer pages • Write the record to the output buffer. If the output buffer is full write it to disk. • Delete the record from its input buffer page.If the buffer page becomes empty then read the next block (if any) of the run into the buffer. • until all input buffer pages are empty:

18. External Sort-Merge (Cont.) • If N M, several merge passes are required. • In each pass, contiguous groups of M - 1 runs are merged. • A pass reduces the number of runs by a factor of M -1, and creates runs longer by the same factor. • E.g. If M=11, and there are 90 runs, one pass reduces the number of runs to 9, each 10 times the size of the initial runs • Repeated passes are performed till all runs have been merged into one.

19. Example: External Sorting Using Sort-Merge M=3; N=4 One tuple in one block

20. External Merge Sort (Cont.) • Cost analysis: • Total number of merge passes required: logM–1(br /M). br /M – nr. of runs after the first step, before merge • Block transfers for initial run creation as well as in each pass is 2br • for final pass, we don’t count write cost • we ignore final write cost for all operations since the output of an operation may be sent to the parent operation without being written to disk • Thus total number of block transfers for external sorting:br ( 2 logM–1(br / M) + 1) • Seeks: next slide

21. External Merge Sort (Cont.) • Cost of seeks • During run generation (first step): one seek to read each run and one seek to write each run • 2 br / M • During the merge phase • Buffer size: bb (read/write bb blocks at a time from each run) • Need 2 br / bb seeks for each merge pass - data is read bb blocks at a time • except the final one which does not require a write • Total number of seeks:2 br / M + br / bb (2 logM–1(br / M) -1)

22. Join Operation • Several different algorithms to implement joins • Nested-loop join • Block nested-loop join • Indexed nested-loop join • Merge-join • Hash-join • Choice based on cost estimate • Examples use the following information • Number of records of customer: 10,000 depositor: 5000 • Number of blocks of customer: 400 depositor: 100

23. Nested-Loop Join • To compute the theta join rsfor each tuple tr in r do begin for each tuple tsin s do begintest pair (tr,ts) tosee if they satisfy the join condition  if they do, add tr• tsto the result.endend • r is called the outerrelation and s the inner relation of the join. • Requires no indices and can be used with any kind of join condition. • Expensive since it examines every pair of tuples in the two relations.

24. Nested-Loop Join (Cont.) • In the worst case, if there is enough memory only to hold one block of each relation, the estimated cost is: • block transfers : nr bs + br ( nr –number of tuples in relation r) (br -number of blocks containing records from relation r) • seeks : nr+ br - one seek for each scan on the inner relation - s is read sequentially - br seeks to read r • If the smaller relation fits entirely in memory, use that as the inner relation. • Reduces cost to br + bsblock transfers and 2 seeks (seeks the first block) • Assuming worst case memory availability cost estimate is • with depositor as outer relation: • 5000  400 + 100 = 2,000,100 block transfers, • 5000 + 100 = 5100 seeks • with customer as the outer relation • 10000  100 + 400 = 1,000,400 block transfers and 10,400 seeks • If smaller relation (depositor) fits entirely in memory, the cost estimate will be 500 block transfers. • Block nested-loops algorithm (next slide) is preferable.

25. Block Nested-Loop Join • Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Brofr do beginfor each block Bsof s do begin for each tuple trin Br do begin for each tuple tsin Bsdo beginCheck if (tr,ts) satisfy the join condition if they do, add tr• tsto the result.end end end end

26. Block Nested-Loop Join (Cont.) • Worst case estimate: br bs + br block transfers + 2 * br seeks • Each block in the inner relation s is read once for each block in the outer relation • 2 seek: after locate one block of the outer, locate one block of the inner, disk arm move, it has to be moved back to the outer relation • Best case: br+ bsblock transfers + 2 seeks. • Improvements to nested loop and block nested loop algorithms: • In block nested-loop, use M — 2 disk blocks as blocking unit for outer relations, where M = memory size in blocks; use remaining two blocks to buffer inner relation and output • Cost = br / (M-2) bs + brblock transfers+ 2br / (M-2) seeks • If equi-join attribute forms a key on the inner relation, stop inner loop on first match • Scan inner loop forward and backward alternately, to make use of the blocks remaining in buffer (with LRU replacement) • Use index on inner relation if available (next slide)

27. Assuming worst case memory availability cost estimate is • with depositor as outer relation: we have to read each block of customer once for each block of depositor • 100  400 + 100 = 40,100 block transfers, • 2 * 100 = 200 seeks • with customer as the outer relation • 400  100 + 400 = 40,400 block transfers and • 2*400=800 seeks • If smaller relation (depositor) fits entirely in memory, the cost estimate will be 500 block transfers too.

28. Indexed Nested-Loop Join • Index lookups can replace file scans if • join is an equi-join or natural join and • an index is available on the inner relation’s join attribute • Can construct an index just to compute a join. • For each tuple trin the outer relation r, use the index to look up tuples in s that satisfy the join condition with tuple tr. • Worst case: buffer has space for only one page of r, and, for each tuple in r, we perform an index lookup on s. • each I/O requires a seek and a block transfer (disk head may have moved between each I/O) • Cost of the join: br(tT + tS) + nr c • Where c is the cost of traversing index and fetching all matching s tuples for one tuple of r • c can be estimated as cost of a single selection on s using the join condition. • If indices are available on join attributes of both r and s,use the relation with fewer tuples as the outer relation.

29. Example of Nested-Loop Join Costs • Compute depositor customer, with depositor as the outer relation. • Let customer have a primary B+-tree index on the join attribute customer-name, which contains 20 entries in each index node. • Since customer has 10,000 tuples, the height of the tree is 4, and one more access is needed to find the actual data • depositor has 5000 tuples • Cost of block nested loops join • 400*100 + 100 = 40,100 block transfers + 2 * 100 = 200 seeks • assuming worst case memory • may be significantly less with more memory • Cost of indexed nested loops join • 100 + 5000 * 5 = 25,100 block transfers and seeks. • CPU cost likely to be less than that for block nested loops join

30. Merge-Join • Sort both relations on their join attribute (if not already sorted on the join attributes). • Merge the sorted relations to join them • Join step is similar to the merge stage of the sort-merge algorithm. • Main difference is handling of duplicate values in join attribute — every pair with same value on join attribute must be matched • Detailed algorithm in book

31. Merge-join algorithm pr:=address of first tuple of r; ps:=address of first tuple of s; while (ps null andpr null) do begin ts := tuple to which ps points; Ss := {tS}; set ps to point to next tuple of s; done := false; while (not done and ps null) do // for one value of the join attributes begin // there can be more tuples ts ':= tuple to which ps points; // in s with the same value if (ts'[JoinAttrs] = ts [JoinAttrs]) // Ss is the set with these then begin // tuples Ss := Ss  {ts '}; set ps to point to next tuple of s; end else done := true; end tr := tuple to which pr points;

32. Merge-join algorithm (cont.) while (pr null and tr [JoinAttrs] < ts [JoinAttrs]) do begin //there aren’t common values in s and r set pr to point to next tuple of r; tr := tuple to which pr points; end while (pr null and tr [JoinAttrs] = ts [JoinAttrs]) do begin for each ts in Ss do begin add ts tr to result end set pr to point to next tuple of r; tr := tuple to which pr points; end end.

33. Merge-Join (Cont.) • Can be used only for equi-joins and natural joins • Each block needs to be read only once (assuming all tuples for any given value of the join attributes fit in memory • Thus the cost of merge join is: br + bs block transfers + br / bb + bs / bb seeks - data is read bb blocks at a time • + the cost of sorting if relations are unsorted. • hybrid merge-join: If one relation is sorted, and the other has a secondary B+-tree index on the join attribute • Merge the sorted relation with the leaf entries of the B+-tree . • Sort the result on the addresses of the unsorted relation’s tuples • Scan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuples • Sequential scan more efficient than random lookup

34. Hash-Join • Applicable for equi-joins and natural joins. • A hash function h is used to partition tuples of both relations • h maps JoinAttrs values to {0, 1, ..., n}, where JoinAttrs denotes the common attributes of r and s used in the natural join. • r0, r1, . . ., rn denote partitions of r tuples • Each tuple tr r is put in partition ri where i = h(tr[JoinAttrs]). • s0,, s1. . ., sn denotes partitions of s tuples • Each tuple tss is put in partition si, where i = h(ts[JoinAttrs]). • Note: In book, ri is denoted as Hri, si is denoted as Hsi and nis denoted as nh.

35. Hash-Join (Cont.)

36. Hash-Join (Cont.) • r tuples in rineed only to be compared with s tuples in si • Need not be compared with s tuples in any other partition,since: • an r tuple and an s tuple that satisfy the join condition will have the same value for the join attributes. • If that value is hashed to some value i, the r tuple has to be in riand the s tuple in si.

37. Hash-Join Algorithm 1. Partition the relation s using hashing function h. When partitioning a relation, one block of memory is reserved as the output buffer for each partition. 2. Partition r similarly. 3. For each i: (a) Load si into memory and build an in-memory hash index on it using the join attribute. This hash index uses a different hash function than the earlier one h. (b) Read the tuples in ri from the disk one by one. For each tuple tr locate each matching tuple tsin si using the in-memory hash index. Output the concatenation of their attributes. The hash-join of r and s is computed as follows. Relation s is called the build input and r is called the probe input.

38. for each tuple ts in sdo begin /* Partition s */ i := h(ts[JoinAttrs]); si:= si {ts}; end for each tuple tr in rdo begin /* Partition r */ i := h(tr[JoinAttrs]); ri:= ri {tr}; end fori := 0 to nhdo begin /* Perform join on each partition */ read si and build an in-memory hash index on it for each tuple tr in rido begin probe the hash index on sito locate all tuples ts such that ts[JoinAttrs] = tr[JoinAttrs] for each matching tuple ts in sido begin add trts to the result end end end

39. Hash-Join algorithm (Cont.) • The value n and the hash function h is chosen such that each si should fit in memory. • Typically n is chosen as bs/M * f where f is a “fudge factor”, typically around 1.2 (bs/ nr. of blocks of s;M- memory size in pages). • The probe relation partitions ri need not fit in memory • it is best to use the smaller relation as the build (inner) relation • Recursive partitioningrequired if number of partitions n is greater than number of pages M of memory. • instead of partitioning n ways, use M – 1 partitions for s • Further partition the M – 1 partitions using a different hash function • Use same partitioning method on r • Rarely required: e.g., recursive partitioning not needed for relations of 1GB or less with memory size of 2MB, with block size of 4KB.

40. Handling of Overflows • Partitioning is said to be skewed if some partitions have significantly more tuples than some others • Hash-table overflow occurs in partition si if si does not fit in memory. Reasons could be • Many tuples in s with same value for join attributes • Bad hash function • Overflow resolution can be done in build phase • Partition si is further partitioned using different hash function. • Partition ri must be similarly partitioned. • Overflow avoidance performs partitioning carefully to avoid overflows during build phase • E.g. partition build relation into many partitions, then combine them • Both approaches fail with large numbers of duplicates • Fallback option: use block nested loops join on overflowed partitions

41. Cost of Hash-Join • If recursive partitioning is not required: cost of hash join is 3(br+ bs) +4  nh block transfers 3: read+write for partitioning + read the partitions - each nh partition could have a partially filled block, for relation r and s , read and write the block 2( br / bb + bs / bb) seeks for partitioning - data is read bb blocks at a time 2 nh seeks for build and probe phases, 1 seek for each partition • If recursive partitioning required: • number of passes required for partitioningbuild relation s is logM–1(bs) – 1 • best to choose the smaller relation as the build relation. • Total cost estimate is: 2(br + bs logM–1(bs) – 1 + br + bs block transfers + 2(br / bb + bs / bb) logM–1(bs) – 1 seeks • If the entire build input can be kept in main memory no partitioning is required • Cost estimate goes down to br +bs.

42. Example of Cost of Hash-Join customer depositor • Assume that memory size is 20 blocks • bdepositor= 100 and bcustomer= 400. • depositor is to be used as build input. Partition it into five partitions, each of size 20 blocks. This partitioning can be done in one pass. • Similarly, partition customer into five partitions, each of size 80. This is also done in one pass. • Therefore total cost, ignoring cost of writing partially filled blocks: • 3(100 + 400) = 1500 block transfers +2( 100/3 + 400/3) = 336 seeks

43. Hybrid Hash–Join • Useful when memory size are relatively large, and the build input is bigger than memory. Memory size >> n + 1 (needed for partitions) n ~ bs/M • Main feature of hybrid hash join: Keep the first partition of the build relation in memory. • E.g. With memory size of 25 blocks, depositor can be partitioned into five partitions, each of size 20 blocks. • Division of memory: • The first partition occupies 20 blocks of memory • 1 block is used for input, and 1 block each for buffering the other 4 partitions. • customer is similarly partitioned into five partitions each of size 80 • the first is used right away for probing, instead of being written out • Cost of 3(80 + 320) + 20 +80 = 1300 block transfers for hybrid hash join, instead of 1500 with plain hash-join. • Hybrid hash-join most useful if M >> bs/MM >>

44. Complex Joins • Join with a conjunctive condition: r 1  2...  ns • Either use nested loops/block nested loops, or • Compute the result of one of the simpler joins r is • final result comprises those tuples in the intermediate result that satisfy the remaining conditions 1 . . .  i –1 i +1 . . .  n • Join with a disjunctive condition r 1  2 ...  ns • Either use nested loops/block nested loops, or • Compute as the union of the records in individual joins r  is: (r 1s)  (r 2s)  . . .  (r ns)

45. Other Operations • Duplicate elimination can be implemented via hashing or sorting. • On sorting duplicates will come adjacent to each other, and all but one set of duplicates can be deleted. • Optimization: duplicates can be deleted during run generation as well as at intermediate merge steps in external sort-merge. • Hashing is similar – duplicates will come into the same bucket. • Projection: • perform projection on each tuple • followed by duplicate elimination.

46. Other Operations : Aggregation • Aggregation can be implemented in a manner similar to duplicate elimination. • Sorting or hashing can be used to bring tuples in the same group together, and then the aggregate functions can be applied on each group. • Optimization: combine tuples in the same group during run generation and intermediate merges, by computing partial aggregate values • For count, min, max, sum: keep aggregate values on tuples found so far in the group. • When combining partial aggregate for count, add up the aggregates • For avg, keep sum and count, and divide sum by count at the end

47. Other Operations : Set Operations • Set operations (,  and ): can either use variant of merge-join after sorting, or variant of hash-join. • E.g., Set operations using hashing: • Partition both relations using the same hash function • Process each partition i as follows. • Using a different hashing function, build an in-memory hash index on ri. • Process si as follows • r  s: • Add tuples in si to the hash index if they are not already in it. • At end of si add the tuples in the hash index to the result. • r  s: • output tuples in sito the result if they are already there in the hash index • r – s: • for each tuple in si, if it is there in the hash index, delete it from the index. • At end of si add remaining tuples in the hash index to the result.

48. Other Operations : Outer Join • Outer join can be computed either as • A join followed by addition of null-padded non-participating tuples. • by modifying the join algorithms. • Modifying merge join to compute r s • In r s, non participating tuples are those in r – R(r s) • Modify merge-join to compute r s: During merging, for every tuple trfrom r that do not match any tuple in s, output tr padded with nulls. • Right outer-join and full outer-join can be computed similarly. • Modifying hash join to compute r s • If r is probe relation, output non-matching r tuples padded with nulls • If r is build relation, when probing keep track of which r tuples matched s tuples. At end of si output non-matched r tuples padded with nulls

49. Evaluation of Expressions • So far: we have seen algorithms for individual operations • Alternatives for evaluating an entire expression tree • Materialization: generate results of an expression whose inputs are relations or are already computed, materialize (store) it on disk. Repeat. • Pipelining: pass on tuples to parent operations even as an operation is being executed • We study above alternatives in more detail

50. Materialization • Materialized evaluation: evaluate one operation at a time, starting at the lowest-level. Use intermediate results materialized into temporary relations to evaluate next-level operations. • E.g., in figure below, compute and storethen compute the store its join with customer, and finally compute the projections on customer-name.